To calculate the change in entropy \( \Delta S^0_{\text{reaction}} \), we can use the relation between the standard electrode potential and the entropy change:
\[
\Delta G^0_{\text{reaction}} = -n F E^0_{\text{cell}}
\]
where:
- \( n = 2 \) (the number of electrons involved in the reaction),
- \( F = 96480 \, \text{C mol}^{-1} \) (Faraday constant),
- \( E^0_{\text{cell}} = 1.38 \, \text{V} \) (standard cell potential).
From the relationship between Gibbs free energy and entropy:
\[
\Delta G^0_{\text{reaction}} = \Delta H^0_{\text{reaction}} - T \Delta S^0_{\text{reaction}}
\]
At constant pressure, \( \Delta H^0_{\text{reaction}} = 0 \) (since the reaction involves no change in enthalpy at standard conditions), so:
\[
\Delta S^0_{\text{reaction}} = - \frac{\Delta G^0_{\text{reaction}}}{T}
\]
Now, using the derivative of the standard potential with respect to temperature \( \left(\frac{\partial E^0}{\partial T}\right)_P = -1.24 \, \text{mV K}^{-1} = -1.24 \times 10^{-3} \, \text{V K}^{-1} \), the entropy change is:
\[
\Delta S^0_{\text{reaction}} = -n F \left(\frac{\partial E^0}{\partial T}\right)_P
\]
Substituting the values:
\[
\Delta S^0_{\text{reaction}} = -2 \times 96480 \times (-1.24 \times 10^{-3})
\]
\[
\Delta S^0_{\text{reaction}} = 239.6 \, \text{J K}^{-1} \, \text{mol}^{-1}
\]
Final Answer:
\[
\boxed{239.6}
\]